asymptotic distribution of joint random variables I am trying to understand the asymptotic distribution of the following expression under normality
$$
{\hat \sigma \hat S - \sigma S}
$$
Where $\sigma$ and $S$ are the population standard deviation and population skewness respectively and $\hat{\sigma}$ and $\hat{S}$ are the sample standard deviation and sample skewness respectively. Any bright ideas, references or suggestions are appreciated
I also need to understand the following covariance structure
$$
{\mathop{\rm cov}} \left[ {\hat \mu  - \mu ,\hat \sigma \hat S - \sigma S} \right]
$$
Where $\mu$ is population mean and $\hat{\mu}$ is the sample mean
What I have is some standard results
$$
\frac{{\sqrt n \left( {\hat \mu  - \mu } \right)}}{\sigma }\xrightarrow{d}{\cal N}\left( {0,1} \right)
$$
$$
\,\frac{{\sqrt n \left( {\hat \sigma  - \sigma } \right)}}{\sigma }\xrightarrow{d}{\cal N}\left( {0,\frac{{\kappa  - 1}}{4}} \right)
$$
$$
\sqrt {\frac{N}{6}} \hat S\xrightarrow{d}{\cal N}\left( {0,1} \right)
$$
 A: I will use the established symbol for skewness, $\gamma_1$. The population Skewness is
$$\gamma_1 = \frac {\mu_3}{\sigma^3} \Rightarrow \sigma \gamma_1 = \frac {\mu_3}{\sigma^2}$$
where $\mu_3$ is the third central moment. For the normal distribution this is zero, so the product $\sigma\gamma_1$ will also be zero.
Assuming an i.i.d. normal sample of size $n$, using the sample analogues of the moments (no need to apply finite sample corrections since we are interested in the asymptotic behavior)
$$\hat \sigma \hat \gamma_1 = \frac {(1/n)\sum_{i=1}^n \left(X_i-\bar X\right)^3}{(1/n)\sum_{i=1}^n \left(X_i-\bar X\right)^2} = \frac {\hat \mu_3}{\hat \sigma^2} \xrightarrow{p} 0$$
Then (see for example Dasgupta 2008, Theorem 3.8, page 42) we have that 
$$\sqrt n\hat \mu_3 \xrightarrow{d} {\cal N}(0, V_3)$$
$$V_3 = \mu_6 - \mu_3^2 -6\sigma^2\mu_4 + 9\sigma^6,\;\; \mu_k = E(X-\mu)^k $$
Using the values and relations between the central moments of the normal distribution we obtain $V_3 = 6\sigma^6$
Since moreover $\hat \sigma^2 \xrightarrow{p}\sigma^2$, we obtain  
$$\sqrt n \hat \sigma \hat \gamma_1 = \frac {\sqrt n\hat \mu_3}{\hat \sigma^2} \xrightarrow{d} {\cal N}(0, 6\sigma^{2})$$
by Slutsky's lemma. Of course the above could be directly derived using again Slutsky's lemma from
$$\sqrt {\frac{n}{6}} \hat \gamma_1 \xrightarrow{d} {\cal N}\left( 0,1 \right) \Rightarrow \sqrt n \hat \sigma \hat \gamma_1 \xrightarrow{d} \sigma \sqrt 6 {\cal N}(0,1) = {\cal N}\left( {0,6\sigma^2}\right)$$
